Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int mm = 100005;
const int mod = 1000000007;
using namespace std;
vector<int> G[mm];
vector<int> c[20][mm];
int par[20][mm];
int d[mm];
int n, m, q;
int lg[mm];
void dfs(int rt) {
  for (int i = 0; 1 << i < d[rt]; i++) {
    int &fa = par[i][rt];
    if (fa) par[i + 1][rt] = par[i][fa];
    vector<int> &vec = c[i + 1][rt];
    vector<int> &vec1 = c[i][rt];
    vector<int> &vec2 = c[i][fa];
    vec.resize(vec1.size() + vec2.size());
    merge(vec1.begin(), vec1.end(), vec2.begin(), vec2.end(), vec.begin());
    if (vec.size() > 10) vec.resize(10);
  }
  for (int i = 0; i < G[rt].size(); i++)
    if (G[rt][i] != par[0][rt]) {
      int &u = G[rt][i];
      par[0][u] = rt;
      d[u] = d[rt] + 1;
      dfs(u);
    }
}
int LCA(int u, int v) {
  while (d[u] != d[v]) {
    if (d[u] > d[v]) swap(u, v);
    int i = lg[d[v] - d[u]];
    v = par[i][v];
  }
  if (u == v) return v;
  for (int t = lg[d[u]]; t >= 0; t) {
    int i = 0;
    while (par[i][u] != par[i][v]) i++;
    if (i == 0) return par[0][u];
    u = par[i - 1][u];
    v = par[i - 1][v];
  }
  return v;
}
vector<int> get(int u, int up, int a, vector<int> &v) {
  vector<int> res[2];
  vector<int> *pre = &res[0], *cur = &res[1];
  while (up > 0) {
    int i = lg[up];
    vector<int> &vec = c[i][u];
    pre->resize(cur->size() + vec.size());
    merge(cur->begin(), cur->end(), vec.begin(), vec.end(), pre->begin());
    if (pre->size() > a) pre->resize(a);
    swap(cur, pre);
    u = par[i][u];
    up -= 1 << i;
  }
  v = *cur;
  return *cur;
}
void solve(int u, int v, int a) {
  int anc = LCA(u, v);
  vector<int> v1;
  get(u, d[u] - d[anc] + 1, a, v1);
  vector<int> v2;
  get(v, d[v] - d[anc], a, v2);
  vector<int> v3(v1.size() + v2.size());
  merge(v1.begin(), v1.end(), v2.begin(), v2.end(), v3.begin());
  a = min(a, (int)v3.size());
  printf("%d", a);
  for (int i = 0; i < a; i++) {
    printf(" %d", v3[i]);
  }
  cout << endl;
}
vector<int> A;
int main() {
  cin >> n >> m >> q;
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    G[x].push_back(y);
    G[y].push_back(x);
  }
  for (int i = 0; i < m; i++) {
    int x;
    scanf("%d", &x);
    c[0][x].push_back(i + 1);
  }
  for (int i = 1; i <= n; i++) {
    sort(c[0][i].begin(), c[0][i].end());
    if (c[0][i].size() > 10) c[0][i].resize(10);
  }
  lg[0] = -1;
  for (int i = 1; i <= n; i++)
    lg[i] = (i & (i - 1)) ? lg[i - 1] : lg[i - 1] + 1;
  int root = 1;
  d[root] = 1;
  dfs(root);
  for (int i = 0; i < q; i++) {
    int u, v, a;
    scanf("%d%d%d", &u, &v, &a);
    solve(u, v, a);
  }
  return 0;
}